Question

Решить уравнение
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Answer 1

$5^{x} \cdot \sqrt[x]{8^{x-1}} = 500$

$5^{x} \cdot 2^{\tfrac{3x - 3}{x} } = 5^{3} \cdot 2^{2}, ~ x \in \mathbb{N}, ~ x > 1$

$\dfrac{2^{\tfrac{3x - 3}{x} }}{2^{2}} = \dfrac{5^{3}}{5^{x}}, ~ x \in \mathbb{N}, ~ x > 1$

$2^{\tfrac{x - 3}{x} } = 5^{3-x}, ~ x \in \mathbb{N}, ~ x > 1$

$\log_{2}2^{\tfrac{x - 3}{x} } = \log_{2}5^{3-x}, ~ x \in \mathbb{N}, ~ x > 1$

$\dfrac{x-3}{x} = (3-x)\log_{2}5, ~ x \in \mathbb{N}, ~ x > 1$

$x - 3 = (3-x)x\log_{2}5, ~ x \in \mathbb{N}, ~ x > 1$

$x - 3 = -(x-3)x\log_{2}5, ~ x \in \mathbb{N}, ~ x > 1$

$(x - 3) + (x-3)x\log_{2}5 = 0, ~ x \in \mathbb{N}, ~ x > 1$

$(x-3)(1 + x\log_{2}5) = 0, ~ x \in \mathbb{N}, ~ x > 1$

$\displaystyle \left [ {{x-3 = 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~} \atop {1 + x\log_{2}5 = 0, ~ x \in \mathbb{N}, ~ x > 1;}} \right.$

$\displaystyle \left [ {{x = 3 \in \mathbb{N},~~~~~~~~} \atop {x = -\dfrac{1}{\log_{2}5} \notin \mathbb{N}.}} \right.$

$x = 3$

Ответ: $x = 3. ~\blacktriangleleft$